A Si sample with 1015/cm3 donors is uniformly optically excited at room temperature such that 1019/cm3 EHPs are generated per second. Find the separation of the quasi-Fermi levels and the change of conductivity upon shining the light. Electron and hole lifetimes are both 10 μs. Dp = 12 cm2/s.
Why did the electron go to the optician?
Because it was feeling a bit "light" headed!
Now, let's tackle your question. To find the separation of the quasi-Fermi levels (ΔE), we can use the equation:
ΔE = kT * ln(Nd / (ni^2 * ne_phi))
Where:
- k is the Boltzmann constant
- T is the temperature
- Nd is the donor concentration
- ni is the intrinsic carrier concentration
- ne_phi is the electron concentration at the quasi-Fermi level
Now, let's calculate it step by step. The intrinsic carrier concentration (ni) can be found using the equation:
ni^2 = np * ne
Where:
- np is the hole concentration
- ne is the electron concentration
- Both np and ne can be found using the equation:
np = ne^2 / ni
Given that the donor concentration (Nd) is 1015/cm3 and the electron-hole pairs generation rate is 1019/cm3 per second, we have all the components to calculate our answer.
But wait, there's more! To find the change in conductivity, we'll need to use the equation:
Δσ = q * (μp * Δp + μn * Δn)
Where:
- Δσ is the change in conductivity
- q is the elementary charge
- μp is the hole mobility
- Δp is the change in the hole concentration
- μn is the electron mobility
- Δn is the change in the electron concentration
Since we were not given the hole mobility or the change in the hole concentration, I'm afraid I can't provide the exact change in conductivity. However, you can use the provided equation along with the given values to calculate it.
I hope this helps shed some light on your question! Let me know if there's anything else I can assist you with.
To find the separation of the quasi-Fermi levels (∆Ef) and the change of conductivity upon shining the light (∆σ), we can follow these steps:
Step 1: Calculate the excess carrier density (∆n) created by optical excitation.
Step 2: Determine the separation of the quasi-Fermi levels (∆Ef).
Step 3: Calculate the change in conductivity (∆σ) due to the optical excitation.
Step 1: Calculate the excess carrier density (∆n):
The excess carrier density ∆n can be calculated using the following equation:
∆n = Gτ
where G is the generation rate and τ is the carrier lifetime.
Given:
G = 1019/cm3 EHPs per second,
τ = 10 μs = 10^-5 s
∆n = (1019/cm3) * (10^-5 s)
∆n = 10^14/cm3
Step 2: Determine the separation of the quasi-Fermi levels (∆Ef):
The separation of the quasi-Fermi levels (∆Ef) can be related to the excess carrier density (∆n) using the following equation:
∆n = (np)^2 / Nc
where np is the excess minority carrier concentration and Nc is the effective density of states in the conduction band.
Given that the sample is uniform, we can assume doping levels are much higher than ∆n. Therefore, np ≈ ∆n. Assuming the sample is intrinsic, Nc is the same as the classical intrinsic carrier concentration (ni).
∆Ef = kT * ln(∆n/ni)
where k is Boltzmann's constant and T is the temperature in Kelvin.
Given:
∆n = 10^14/cm3
ni ≈ 1.5 * 10^10/cm3 at room temperature (approximation)
∆Ef = (1.38 * 10^-23 J/K) * (300 K) * ln(10^14/1.5 * 10^10)
∆Ef ≈ 0.285 eV
Step 3: Calculate the change in conductivity (∆σ):
The change in conductivity (∆σ) is related to the change in electron concentration (∆n) using the following equation:
∆σ = qμn * ∆n
where q is the charge of an electron and μn is the electron mobility.
Given:
Dp = 12 cm2/s (diffusion coefficient)
Dp = kT * μp / q (Einstein relation, assuming μp ≈ μn)
∆σ = q * ∆n * μp
∆σ = q * ∆n * (Dp * q / kT)
Using q = 1.6 * 10^-19 C and k = 1.38 * 10^-23 J/K:
∆σ = (1.6 * 10^-19 C) * (10^14/cm3) * (12 cm2/s) * (1.6 * 10^-19 C / (1.38 * 10^-23 J/K) * (300 K))
∆σ ≈ 11.63 Ω^-1 * cm^-1
Therefore, the separation of the quasi-Fermi levels (∆Ef) is approximately 0.285 eV, and the change in conductivity (∆σ) upon shining the light is approximately 11.63 Ω^-1 * cm^-1.
To find the separation of the quasi-Fermi levels and the change in conductivity upon shining light on a Si sample with given parameters, we need to follow these steps:
Step 1: Calculate the excess carrier concentration.
Given that the number of electron-hole pairs (EHPs) generated per second is 1019/cm3, and both electrons and holes have a lifetime of 10 μs, we can calculate the excess carrier concentration as follows:
Excess carrier concentration = EHPs generated per second * Lifetime of carriers
= (1019/cm3) * (10 μs)
= 1010 cm-3
Step 2: Calculate the change in electron and hole concentrations.
Since the sample is initially doped with 1015/cm3 donors, we assume that the intrinsic carrier concentration ni can be neglected compared to the doping concentration.
Therefore, the change in electron and hole concentrations is equal to the excess carrier concentration:
Δn = Δp = excess carrier concentration = 1010 cm-3
Step 3: Calculate the change in electron and hole mobilities.
Given the hole diffusion constant Dp as 12 cm2/s, we can calculate the hole mobility μp using the Einstein relationship:
μp = Dp / (kT/q)
= (12 cm2/s) / (1.38 x 10^-23 J/K * 300 K / 1.6 x 10^-19 C)
= 0.005 cm2/Vs
Assuming the electron mobility is the same as the hole mobility:
μn = μp = 0.005 cm2/Vs
Step 4: Calculate the change in conductivity (Δσ).
Using the change in carrier concentration and mobility, we can calculate the change in conductivity as follows:
Δσ = (q * Δp * μp) + (q * Δn * μn)
= (1.6 x 10^-19 C) * (1010 cm-3) * (0.005 cm2/Vs) + (1.6 x 10^-19 C) * (1010 cm-3) * (0.005 cm2/Vs)
= 0.16 (1/Ωm)
Step 5: Calculate the separation of the quasi-Fermi levels (ΔEf).
Given that ΔEf = kT * ln(Δn/ni), where ni is the intrinsic carrier concentration, we can calculate the separation of the quasi-Fermi levels as follows:
ΔEf = (1.38 x 10^-23 J/K) * (300 K) * ln[(1010 cm-3)/ni]
Since the sample is heavily doped (1015/cm3), we can neglect the contribution of the intrinsic carrier concentration, and thus ΔEf ≈ 0.
Therefore, the separation of the quasi-Fermi levels is negligible in this case.
In summary:
- The separation of the quasi-Fermi levels (ΔEf) is negligible due to heavy doping.
- The change in conductivity upon shining light (Δσ) is 0.16 (1/Ωm).